JEE Main 2026 (Online) 2nd April Morning ShiftQuestion 1
The number of elements in the set $S=\left\{(r,k):k\in\mathbb{Z}\text{ and }{}^{36}C_{r+1}=\frac{6\left({}^{35}C_{r}\right)}{(k^{2}-3)}\right\}$, is :
Permutation and Combination
Solution
Concept. Use the ratio ${}^{36}C_{r+1}/{}^{35}C_r$ to turn the condition into a simple equation in $r$ and $k$. Formulas used. (1) ${}^{n}C_{r}=\dfrac nr\,{}^{n-1}C_{r-1}$, so ${}^{36}C_{r+1}=\dfrac{36}{r+1}\,{}^{35}C_r$Step 1 - Ratio. The condition becomes $$\frac{36}{r+1}=\frac{6}{k^2-3}\ \Rightarrow\ r+1=6(k^2-3)$$ Step 2 - Range. $0\le r\le35$, so $1\le r+1\le36$. Hence $1\le k^2-3\le6$, i.e. $4\le k^2\le9$. Step 3 - Integer $k$. $k^2=4$ or $9$ (5, 6, 7, 8 are not squares). $k^2=4$: $r+1=6$, $r=5$, $k=\pm2$. $k^2=9$: $r+1=36$, $r=35$, $k=\pm3$. Step 4 - Count. $(5,2),(5,-2),(35,3),(35,-3)$: $4$ elements. Answer: B